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  • Department of Mechanical Engineering Prepared By: Vimal Limbasiya Darshan Institute of Engineering & Technology, Rajkot Page 1.1

    1 BALANCING OF ROTATING MASSES

    Course Contents

    1.1 Introduction

    1.2 Static Balancing

    1.3 Types of Balancing

    1.4 Balancing of Several Masses

    Rotating in the Same Plane

    1.5 Dynamic Balancing

    1.6 Balancing of Several Masses

    Rotating in the different

    Planes

    1.7 Balancing Machines

  • 1. Balancing of Rotating Masses Dynamics of Machinery (2161901)

    Prepared By: Vimal Limbasiya Department of Mechanical Engineering Page 1.2 Darshan Institute of Engineering & Technology, Rajkot

    1.1 Introduction

     Often an unbalance of forces is produced in rotary or reciprocating machinery due

    to the inertia forces associated with the moving masses. Balancing is the process of

    designing or modifying machinery so that the unbalance is reduced to an

    acceptable level and if possible is eliminated entirely.

    Fig. 1.1

     A particle or mass moving in a circular path experiences a centripetal acceleration

    and a force is required to produce it. An equal and opposite force acting radially

    outwards acts on the axis of rotation and is known as centrifugal force [Fig. 1.1(a)].

    This is a disturbing force on the axis of rotation, the magnitude of which is constant

    but the direction changes with the rotation of the mass.

     In a revolving rotor, the centrifugal force remains balanced as long as the centre of

    the mass of the rotor lies on the axis of the shaft. When the centre of mass does

    not lie on the axis or there is an eccentricity, an unbalanced force is produced [Fig.

    1.1(b)]. This type of unbalance is very common. For example, in steam turbine

    rotors, engine crankshafts, rotary compressors and centrifugal pumps.

     Most of the serious problems encountered in high-speed machinery are the direct

    result of unbalanced forces. These forces exerted on the frame by the moving

    machine members are time varying, impart vibratory motion to the frame and

    produce noise. Also, there are human discomfort and detrimental effects on the

    machine performance and the structural integrity of the machine foundation.

     The most common approach to balancing is by redistributing the mass which may

    be accomplished by addition or removal of mass from various machine members.

     There are two basic types of unbalance-rotating unbalance and reciprocating

    unbalance – which may occur separately or in combination.

    1.2 Static Balancing:

     A system of rotating masses is said to be in static balance if the combined mass

    centre of the system lies on the axis of rotation.

  • Dynamics of Machinery (2161901) 1. Balancing of Rotating Masses

    Department of Mechanical Engineering Prepared By: Vimal Limbasiya Darshan Institute of Engineering & Technology, Rajkot Page 1.3

    1.3 Types of Balancing:

    There are main two types of balancing conditions

    (i) Balancing of rotating masses

    (ii) Balancing of reciprocating masses

    (i) Balancing of Rotating Masses

    Whenever a certain mass is attached to a rotating shaft, it exerts some centrifugal

    force, whose effect is to bend the shaft and to producevibrations in it. In order to prevent

    the effect of centrifugal force, another mass is attached to the opposite side of theshaft,

    at such a position so as to balance the effect of the centrifugal force of the first mass.

    This is done in such away that the centrifugal forces of both the masses are made to be

    equal and opposite. The process of providing the second mass in order to counteract the

    effect of the centrifugal force of the first mass is called balancing of rotating masses.

    The following cases are important from the subject point of view:

    1. Balancing of a single rotating mass by a single mass rotating in the same plane.

    2. Balancing of different masses rotating in the same plane.

    3. Balancing of different masses rotating in different planes.

    1.4 Balancing of Several Masses Rotating in the Same Plane

     Consider any number of masses (say four) of magnitude m1, m2, m3 and m4 at

    distances ofr1, r2, r3 and r4 from the axis of the rotating shaft. Let 1, 2,3and 4be

    the angles of these masses with the horizontal line OX, as shown in Fig. 1.2 (a). Let

    these masses rotate about an axis through O and perpendicular to the plane of

    paper, with a constant angular velocity of  rad/s.

    (a)Space diagram. (b) Vector diagram.

    Fig. 1.2 Balancing of several masses rotating in the same plane.

     The magnitude and position of the balancing mass may be found out analytically

    or graphically as discussed below:

    1. Analytical method

  • 1. Balancing of Rotating Masses Dynamics of Machinery (2161901)

    Prepared By: Vimal Limbasiya Department of Mechanical Engineering Page 1.4 Darshan Institute of Engineering & Technology, Rajkot

     Each mass produces a centrifugal force acting radially outwards from the axis of

    rotation. Let F be the vector sum of these forces.

    F = m1r1 2+ m2r2

    2 +m3r3 2+ m4r4

    2

     The rotor is said to be statically balanced if the vector sum F is zero.

     If F is not zero, i.e., the rotor is unbalanced, then produce a counterweight

    (balance weight) of mass mc, at radius rc to balance the rotor so that

    m1r1 2+ m2r2

    2 +m3r3 2+ m4r4

    2 + mcrc 2 = 0

    m1r1+ m2r2 +m3r3+ m4r4 + mcrc = 0

     The magnitude of either mc or rc may be selected and of other can be calculated.

     In general, if mr is the vector sum of m1.r1, m2.r2, m3.r3, m4.r4, etc., then

    mr +mcrc = 0

     To solve these equation by mathematically, divide each force into its x and z

    components, mrcos + mcrccosc = 0

    and mrsin +mcrcsinc= 0

    mcrccosc= −mrcos…………………………(i)

    and mcrcsinc= −mrsin............................(ii)

     Squaring and adding (i) and (ii),

    mcrc = 𝑚𝑟𝑐𝑜𝑠 ² + 𝑚𝑟𝑠𝑖𝑛 ²

     Dividing (ii) by (i),

    𝑡𝑎𝑛𝑐 = −𝑚𝑟𝑠𝑖𝑛

    −𝑚𝑟𝑐𝑜𝑠

     The signs of the numerator and denominator of this function identify the quadrant

    of the angle.

    2. Graphical method

     First of all, draw the space diagram with the positions of the several masses, as

    shown in Fig. 1.2 (a).

     Find out the centrifugal force (or product of the mass and radius of rotation)

    exerted byeach mass on the rotating shaft.

     Now draw the vector diagram with the obtained centrifugal forces (or the product

    of themasses and their radii of rotation), such that ab represents the centrifugal

    force exerted by the mass m1 (or m1.r1) in magnitude and direction to some

    suitable scale. Similarly, draw bc, cd and de to represent centrifugal forces of

    other masses m2, m3 and m4 (or m2.r2,m3.r3 and m4.r4).

     Now, as per polygon law of forces, the closing side ae represents the resultant

    force inmagnitude and direction, as shown in Fig. 1.2 (b).

     The balancing force is, then, equal to resultant force, but in opposite direction.

  • Dynamics of Machinery (2161901) 1. Balancing of Rotating Masses

    Department of Mechanical Engineering Prepared By: Vimal Limbasiya Darshan Institute of Engineering & Technology, Rajkot Page 1.5

     Now find out the magnitude of the balancing mass (m) at a given radius of

    rotation (r), such that

    m.r.2= Resultant centrifugal force

    or m.r = Resultant of m1.r1, m2.r2, m3.r3 and m4.r4

     (In general for graphical solution, vectors m1.r1, m2.r2, m3.r3, m4.r4, etc., are added. If

    they close in a loop, the system is balanced. Otherwise, the closing vector will be

    giving mc.rc. Its direction identifies the angular position of the countermass relative

    to the other mass.)

    Example 1.1 :A circular disc mounted on a shaft carries three attached masses of 4 kg, 3 kg

    and 2.5 kg at radial distances of 75 mm, 85 mm and 50 mm and at the angular positions of

    45°, 135° and 240° respectively. The angular positions are measured counterclockwise from

    the reference line along the x-axis. Determine the amount of the countermass at a radial

    distance of 75 mm required for the static balance.

    m1 = 4 kg r1 = 75 mm 1 = 45°

    m2 = 3 kg r2= 85 mm 2 = 135°

    m3 = 2.5 kg r3 = 50 mm 3 = 240°

    m1r1 = 4 x 75 = 300 kg.mm

    m2r2 = 3 x 85 = 255 kg.mm

    m3r3 = 2.5 x 50 = 125 kg.mm

    Analytical Method:

    mr +mcrc = 0

    300 cos45°+ 255 cos135° + 125 cos240° + mcrccosc= 0 and

    300 sin 45°+ 255 sin 135° + 125 sin 240° + mcrcsinc= 0

    Squaring, adding and then solving,

    2

    2

    (300 cos 45 255 cos 135 125 cos240 )

    (300 sin45 255 sin 135 125 sin240 ) C Cm r

         

       

    2 275